Lösning 1.9.4a
Förberedande kurs i matematik
(Skillnad mellan versioner)
(Ny sida: Sant, eftersom <math>\text{Re}(z)=\text{Re}(a+bi)=a</math> <math>\text{Re}(\bar{z})=\text{Re}(a-bi)=a</math>) |
|||
(En mellanliggande version visas inte.) | |||
Rad 1: | Rad 1: | ||
- | Sant, eftersom <math>\text{Re}(z)=\text{Re}(a+bi)=a</math> | + | Sant, eftersom |
+ | |||
+ | <math>\text{Re}(z)=\text{Re}(a+bi)=a</math> | ||
+ | |||
<math>\text{Re}(\bar{z})=\text{Re}(a-bi)=a</math> | <math>\text{Re}(\bar{z})=\text{Re}(a-bi)=a</math> |
Nuvarande version
Sant, eftersom
\displaystyle \text{Re}(z)=\text{Re}(a+bi)=a
\displaystyle \text{Re}(\bar{z})=\text{Re}(a-bi)=a